Find the point on the curve …
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Tripti Rawat 6 years, 4 months ago
The given curve is y=[x(x-2)]{tex}^2{/tex}
Then, y=[x{tex}^2{/tex}-2x]{tex}^2{/tex}
{tex} \frac { d y } { d x } = 0{/tex}
{tex} \Rightarrow \quad \frac { d } { d x } \left( x ^ { 2 } - 2 x \right) ^ { 2 } = 0{/tex}
{tex} \Rightarrow{/tex}2(x2 - 2x)(2x - 2) = 0
{tex} \Rightarrow{/tex} x = 0, 1, 2
When x = 0, then y = [0 -(-2)]2 = 0
When x = 1, then y = [1 - 2(1)]2 = 1
When x = 2, then y = [22 - 2 {tex} \times{/tex} 2]2 = 0
Hence, the tangent is parallel to X-axis at the points (0, 0), ( 1, 1) and (2, 0).
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